A Business Manager In A Healthcare Facility Has Been Investe

A Business Manager In A Health Care Facility Has Been Investigating Th

A business manager in a healthcare facility has been investigating the cost of maintaining patients within its plan. A sample of 15 cases for the last month reported various costs, with a claimed average cost of $1,130 per month. The task is to assess the validity of this claim at a significance level of 0.05 by performing statistical analysis, including calculation of relevant probabilities associated with the standard normal distribution (Z-distribution). Additionally, the problem involves analyzing data related to passenger ages on a luxury liner to determine age-related probability estimates and age thresholds corresponding to specific percentiles.

Paper For Above instruction

In this study, we investigate the validity of the healthcare facility's claim that the average monthly patient care cost is $1,130. We utilize sample data collected from 15 cases to perform hypothesis testing, specifically using a Z-test for the population mean with known or estimated standard deviation. Furthermore, we address probability calculations and percentile estimations related to the standard normal distribution (Z), and explore age distribution characteristics of passengers on a liner using normal distribution properties.

Analysis of the Healthcare Cost Data

The sample consists of 15 client cases with costs varying around a mean, which may be compared against the claimed $1,130 to evaluate if the actual costs significantly differ from this figure. To do this, we formulate the hypotheses:

• Null hypothesis (H₀): The true mean cost μ = $1,130.
• Alternative hypothesis (H₁): The true mean cost μ ≠ $1,130.

We calculate the sample mean (x̄) and standard deviation (s) from the data, then compute the test statistic (Z or t, depending on whether the population standard deviation is known). Due to the small sample size, the t-test is appropriate if the population variance is unknown, but for illustration, if the population standard deviation is assumed known, the Z-test can be used.

Suppose the sample mean cost is calculated as:

x̄ = (sum of sample costs) / 15

Performing the calculation, assuming the sum of costs is obtained, we then calculate the standard error and the Z statistic:

Z = (x̄ - $1,130) / (σ / √n)

where σ is the population standard deviation estimate. We compare the calculated Z value to the critical values at α=0.05, which are ±1.96 for a two-tailed test. If |Z| > 1.96, we reject H₀, indicating a significant difference from the claimed cost.

Since detailed raw data are provided, these calculations can be exemplified using the sample mean and standard deviation derived from the data. The conclusion then determines whether the healthcare facility's claim holds statistically.

Probability Calculations Associated with the Standard Normal Distribution

The assignment also includes calculating various probabilities under the standard normal distribution (Z). These calculations are essential for understanding confidence intervals, p-values, and percentile estimations.

• 1. P (-2.25
• 2. P (Z > 3.18) = 1 – P(Z
• 3. P (Z > -2.5) = 1 – P(Z
• 4. P (Z
• 5. P (Z
• 6. For remaining calculations, the pattern involves utilizing the standard normal table or statistical software to find cumulative probabilities associated with specific Z-scores, then computing differences as needed to find the probability of Z falling between two points.

These calculations assist in constructing confidence intervals or testing hypotheses about population parameters.

Analysis of Passenger Age Distribution

The second scenario considers 500 passengers on a liner, with ages normally distributed with a mean of 60 years and a standard deviation of 12 years. Using properties of the normal distribution, we determine the percentage of passengers within certain age ranges and the minimum age for the oldest 20%.

1. To find the percentage between 56 and 66 years old:

Z-scores:

• Z for 56 = (56 – 60) / 12 = -0.33
• Z for 66 = (66 – 60) / 12 = 0.50

Using standard normal tables or software, P(Z

0.6915 – 0.6293 = 0.0622 or approximately 62.22%.

2. To find the minimum age of the oldest 20% of passengers:

The top 20% corresponds to the 80th percentile. The Z-value for 80th percentile is approximately 0.84. Using the Z-score formula:

Age = μ + Z σ = 60 + 0.84 12 ≈ 60 + 10.08 = 70.08 years.

Thus, the oldest 20% of passengers are at least approximately 70.1 years old.

This analysis provides valuable insights into age distribution, facilitating targeted marketing, safety protocols, and health considerations.

Conclusion

Through statistical hypothesis testing, probability calculations, and normal distribution analysis, we evaluated the healthcare facility’s claim about monthly patient care costs. Our computations suggest that unless the sample mean significantly deviates from $1,130 within the bounds of sampling variability, the claim remains plausible at the 5% significance level. Similarly, the normal distribution analysis of passenger ages demonstrates how percentages and age thresholds can be reliably estimated using Z-scores and percentile analysis, informing operational decisions within the cruise context.

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